$\mathbb{CP}^1$ is the set of all one dimensional subspaces of $\mathbb{C}^2$, if $(z,w)\in \mathbb{C}^2$ be non zero , then its span is a point in $\mathbb{CP}^1$.let $U_0=\{[z:w]:z\neq 0\}$ and $U_1=\{[z:w]:w\neq 0\}$, $(z,w)\in \mathbb{C}^2$,and $[z:w]=[\lambda z:\lambda w],\lambda\in\mathbb{C}^{*}$ is a point in $\mathbb{CP}^1$, the map is $\phi_0:U_0\rightarrow\mathbb{C}$ defined by $$\phi_0([z:w])=w/z$$
the map $\phi:U_1\rightarrow\mathbb{C}$ defined by $$\phi_1([z:w])=z/w$$
Now,Could any one tell me why $\mathbb{CP}^1$ is the union of two closed sets $\phi_0^{-1}(D)$ and $\phi_1^{-1}(D)$, where $D$ is closed unit disk in $\mathbb{C}$,and why $\mathbb{CP}^1$ is compact?

What is the topology over $\mathbb{CP}^1$?
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SeiriosJul 25 '12 at 9:31

Dear Patience, seeing that you have asked 11 questions (about Riemann surfaces) in two says, you might like to know that there is a limit of 50 questions you can ask per month: see here
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wildildildlifeJul 25 '12 at 12:10

5 Answers
5

The maps you gave are the coordinate charts on $\mathbb{C}\mathbb{P}^1$ that makes it into a manifold. In particular, they are bijective.

If we take $\phi_0^{-1}(D)$, we get all point $[1:z]$, with $z\in D$. Similarly, $\phi_0^{-1}(D)$ is all points of the form $[z:1]$. Together, these sets cover $\mathbb{C}\mathbb{P}^1$. The inverse maps are continuous, because the maps $\phi_i$ are bijective, so $\phi_0^{-1}(D)$ and $\phi_1^{-1}(D)$ are compact, as the image of a compact set under a continuous map is compact. It's not hard to show that if a space is a union of two compact sets, it is compact, so we are done.

Ok, nevermind, this is right. To answer your question: it doesn't make sense to speak of coordinate charts being holomophic bijections. You want the transition functions to be holomorphic.
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PotatoJul 25 '12 at 9:47

So you had two questions, one about the charts covering the space, and one about compactness.

The first has been answered, but i would like to add some details.
So you do see that the $U_i = \phi_i^{-1}(\mathbb{C})$ for $i=0,1$ cover $\mathbb{C} P^1$ right? My claim is now that we can restrict to the closed disks in $\mathbb{C}$. The cases $[0:1]$ and $[1:0]$ are simple and the only points outside $U_{0,1} := U_0 \cap U_1$, so we restict to $[z:w] \in U_{0,1}$. Now both $\phi_i$ are defined at this point, and since $\phi_0([z:w]) = \frac{1}{\phi_1([z:w])}$ we can pick the one that has norm smaller or equal to one, so that the image lies in the unit disk. Since the point was arbitrary, every point has as $i$ such that its $\phi_i$ image lies in the unit disk, q.e.d.

Regarding the second one, recall that the continuous image of a compact space is compact. Moreover it is well known that the sphere $S \subset \mathbb{C}^2$ is compact, and that $\mathbb{C}P^1$ is a quotient of this sphere. Summarizing, there is a surjective quotient map $\pi: S \rightarrow \mathbb{C}P^1 $, hence $\mathbb{C}P^1$ is the image of a compact space, hence compact.

The reason i add the second answer is that it is insightful and easily generalized to complex or real projective spaces of any dimension, even to Grassmanians (although that requires some adjustments) and i believe also you can replace $\mathbb{C}$ by any normed division algebra over $\mathbb{R}$.

By the way, note that the spheres are not complex submanifolds of $\mathbb{C}^n$, we just regard them as topological spaces here. Since compactness is purely topological this suffices.
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JoachimJul 25 '12 at 10:01

thank you for the detail and informative answer. simply pleased and delighted
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La Belle NoiseuseJul 25 '12 at 10:01